Home
Class 11
MATHS
Find the equation to the locus of points...

Find the equation to the locus of points equidistant from the points
(a + b,a - b),(a - b,a + b)

Text Solution

Verified by Experts

The correct Answer is:
`x-y=0`
Promotional Banner

Topper's Solved these Questions

  • LOCUS

    AAKASH SERIES|Exercise EXERCISE - 2.1 ( SHORT ANSWER QUESTIONS)|14 Videos

  • LOCUS

    AAKASH SERIES|Exercise ADVANCED SUBJECTIVE TYPE QUESTIONS)|10 Videos

  • LOCUS

    AAKASH SERIES|Exercise PRACTICE EXERCISE|37 Videos

  • LIMITS

    AAKASH SERIES|Exercise PRACTICE EXERCISE|143 Videos

  • MATHEMATICAL INDUCTION

    AAKASH SERIES|Exercise PRACTICE SHEET (EXERCISE-I) LEVEL-I (Principle of Mathematical Induction) (Straight Objective Type Questions)|55 Videos

Similar Questions

Explore conceptually related problems

The equation to the locus of points equidistant from the points ( 2 , 3 ) , ( - 2, 5 ) is

Find the equation to the locus of points equidistant from the points (-3,2),(0,4)

The equation to the locus of points equidistant from the points (-3, 4),(3, 4) is

Find the equation to the locus of points equidistant from the points (i) (-3,2), (0,4) (ii) (a+b,a-b),(a-b,a+b)

If the equation of the locus of points equidistant from the points ( - 2, 3 ) , ( 6, -5 ) is a x + by + c = 0 then ascending order of a, b , c is

If the equation of the locus of a point equidistant from the points ( a_ 1 , b _ 1 ) and ( a _ 2, b _ 2 ) is ( a _ 1- a _ 2 ) x + ( b _ 1 - b _ 2 ) y + c = 0 then the value of c is

A: The equation to the locus of points which are equidistant from the points ( - 3, 2 ) , (0, 4 ) is 6x + 4y - 3 =0 . R : The locus of points which are equidistant to A, B is perpendicular bisector of AB

If the equation to the locus of points equidistant from the points (-2, 3), (6, -5) is ax+by+c=0 where a lt 0 then, the ascending order of a, b, c is

If the equation to the locus of points equidistant from the points (-2,3), (6,-5) is ax+by+c=0 where agt0 then, the ascending order of a, b, c is